Problem: Point $P$ lies on the line $x= -3$ and is 10 units from the point $(5,2)$. Find the product of all possible $y$-coordinates that satisfy the given conditions.
All points on the line $x=-3$ are of the form $(-3,y)$, where $y$ is a real number.  The distance from $(5,2)$ to $(-3,y)$ is $$\sqrt{(5-(-3))^2+(2-y)^2}$$ units.  Setting this expression equal to 10, we find \begin{align*}
\sqrt{(5-(-3))^2+(2-y)^2}&= 10 \\
64+(2-y)^2&= 100 \\
(2-y)^2&= 36 \\
2-y&=\pm 6 \\
y=2\pm6.
\end{align*} The product of $2+6 = 8$ and $2-6 = -4$ is $\boxed{-32}$. [asy]

import graph;

size(200);

defaultpen(linewidth(0.7)+fontsize(10));

dotfactor=4;

xaxis(xmax=7,Ticks(" ",1.0,begin=false,end=false,NoZero,Size=3),Arrows(4));

yaxis(Ticks(" ",1.0,begin=false,end=false,NoZero,Size=3),Arrows(4));

pair A=(5,2), B=(-3,8), C=(-3,-4);

pair[] dots={A,B,C};

dot(dots);

label("(5,2)",A,E);

draw((-3,-6)--(-3,10),linetype("3 3"),Arrows(4));

draw(B--A--C);

label("10",(A+B)/2,NE);

label("10",(A+C)/2,SE);

label("$x=-3$",(-3,-6),S);[/asy]